Current draw of a motor, under different situations

[FOIWATER]

counter emf pertains to all electric motors.

resistance will not change in AC or DC motors, all dependent on CEMF

 

[FOIWATER]

In fact, AC motors will draw up to 6 times full load amperage at starting.

They are where soft starting becomes crucial in industry

 

[psparky]

In fact, AC motors will draw up to 6 times full load amperage at starting.

They are where soft starting becomes crucial in industry

Very aware of this as breakers I size are typically way, way larger than full load amps. A G.E. or square D chart is great for this...or if there actually is a MOCP number available I will use that.

I'll take a look later when I have ample time and try to digest what you said...thanks.

 

[psparky]

Wait...it just clicked...I think I know what you mean.

In a normal magetic field...the current is driving the magnetic field.

Now as soon as the magnetic field changes speed due to exterior forces or perhaps start up or whatever...now all the sudden the current has a counter effect on it from the change in the existing magnetic field...now the magnetic field is now driving or resisting the current that was originally driving it...

And in the case of start up...there is no back EMF...no rotating magnetic field...nothing...just bare wire...hence full load locked rotor current on start up...for a few cycles anyway.

In the ball park?

 

[FOIWATER]

let me try my best.

You have a DC shunt wound motor, very basic..

assume the shunt and armature are fed by two separate sources.

you can think of your shunt and armature as two separate coils in very close proximity physically.

Your armature has a very low resistance about half an ohm. Your shunt has a little higher resistance, between 1 and 10 ohms, depending on the motor.

you connect your motor to a power source.

the motor accepts current through its winding, both the field and the armature based on their resistance. At this point, the value for accepted current is at a maximum value, supply voltage/ coil resistance = current drawn.

The current through the coil develops a magnetic field around the coil.

We now, due to lorentz' force law, have the basic necessities for motoring. we have a current carrying conductor places in a magnetic field, it will now experience torque.

As the conductor begins to move now however, we have the conditions met for GENERATION... based on Faraday induction law.. which states that relative motion between a conductor and a magnetic field will induce a voltage in said conductor. The conductor my friend, is the armature.. and the voltage is the counter electromotive force. Its value is dependent on the value of field flux, and the speed of the motor rotor (dphi/dt).

The counter electromotive force that is induced, causes a current to flow in the armature, which due to lenz' law, OPPOSES the current direction which induced the voltage, that being the SOURCE VOLTAGE.

Therefore, the faster we spin the rotor, the more counter electromotive force is induced in the motors armature. and therefore, the more it opposes the source voltage, IE: less voltage!

Now look at ohms law...

V = IR

It has been said here that for the current to change under load (speed), the resistance MUST be changing. try my friend to imagine that the VOLTAGE is changing with speed (load), not the resistance. The voltage changes because of CEMF---> and this new value of voltage yields a new value for current drawn. for a fixed resistance and supply voltage.

higher speed = more counter emf induced = more bucking cemf = less current drawn.

So, at high speeds... the motor draws little current.

Now, add a load, and people say, "now the motor draws more current, it is trying harder to turn the load!" ... no.

It draws more current, because it slowed down, and as it slowed down, the armature coil was cut by the magnetic lines of flux of the field less over time, and hence a lower cemf was induced in the armature, meaning MORE of the source voltage was received by the motor, meaning MORE current was drawn.

This is as good as I can explain it, sorry

 

[jim hardy]

Simplified equivalent circuit is armature resistance in series with ideal voltage source Ec, counter-emf.

If external voltage < Ec current flows out and it's a generator, if > Ec current flows in and it's a motor.

If Ec = external voltage no current flows, hence no torque.

Any help, guys?

 

[FOIWATER]

yes, ha

much easier way of thinking logically about it, Jim.

 

[saskcat]

hi again. in regard to overloading:
this is known as the breakaway torque. the ac squirrel cage induction motor has markings on it that represent the characteristics of the unit. let's say 100hp 440v 60hz 1745rpm 1.15sf - just for argument sake. i don't know the actual current draw at full load - but not important for this. let's assume it is rated for 125amps. at rated load -torque- the motor will draw 125amps. it will be running at 1745 rpm. with less load, it spin run faster. there r 4 poles developed in that unit and at 60hz the magnetic field rotates around the stator at ~1800rpm.
the rotor can not reach that speed as it needs to cut the magnetic field in order to induce an voltage -field- about itself. now I am not sure whether it is a push pull effect but the rotor chases the field around in a circle. due to the strength of the field built under no load, less field less current. as load increases, rotor slows down, cutting more field, increasing in magnitude and producing more torque. this happens until motor design. full load, full torque, full amps, rated voltage and power factor. after that-if we apply more load, the motor will draw more than its designed rating - it will continue to draw more current - building more field being cut by more flux until it gets saturated and breaks away from the field and stalls. at that point up to 6x rated current is drawn and of coarse-melt down.
there are numbers to represent each motor as each are different by manufacturer, design, size, ect.
im just a sparky but that is the nuts n bolts of it. cheers

 

[Windadct]

The Starting current is high and a High Efficiency motor - is worse, up to 10 X. IN addition to soft starting - a drive will typically pay for itself quickly - since each application does not need the exact nameplate rating of the motor ( for example a 4 hp application needs a 5 HP motor - because 4 HP motors are not common). In this case the motor still draws nearly full current, with a poor power factor - by using a drive it regulates the voltage to make the motor operate at the 4 HP point, better PF, easier starting etc..

Starting a larger 3 PH PM motor - is pretty much impossible without a drive - I was just working with a customer on this : 61KW, 360V, I max 250 A - BUT the Rk(windings) is 0.036 Ohms!
Typical drives would start a PM motor at 50% duty cycle ( like a DC motor) - on this I think we will need 5 -10% duty, at least until we see some back EMF, and then ramp the Duty Cycle up until we see a 1 or 2 rpm then transition the drive the 2 PH AC (Sine) mode. Add to this that in the application the start could be under no load or locked rotor ( un-startable).